Exercise 1.6. Alyssa P. Hacker doesn’t see why if needs to be provided as a special form. “Why can’t I just define it as an ordinary procedure in terms of cond?” she asks. Alyssa’s friend Eva Lu Ator claims this can indeed be done, and she defines a new version of if:

(define (new-if predicate then-clause else-clause)
  (cond (predicate then-clause)
        (else else-clause)))

Eva demonstrates the program for Alyssa:

(new-if (= 2 3) 0 5)

(new-if (= 1 1) 0 5)

Delighted, Alyssa uses new-if to rewrite the square-root program:

(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
          (sqrt-iter (improve guess x)

What happens when Alyssa attempts to use this to compute square roots? Explain.

When Alyssa uses new-if in the sqrt-iter procedure, it gets stuck in an infinite loop.

Because Scheme interpreters use applicative order evaluation, when new-if is called for the first time, the interpreter tries to evaluate all of its arguments. One of its arguments is (sqrt-iter (improve guess x) x), so sqrt-iter is called again, with different arguments. Nothing stops sqrt-iter from being called over and over in an infinite loop, even when the guess is good enough, because the interpreter tries to evaluate sqrt-iter (improve guess x) x) every time.

If the interpreter was using normal order evaluation, it would not get stuck in an infinite loop. When it got to an iteration where the good-enough? test was #t, it wouldn’t try to evaluate (sqrt-iter (improve guess x) x) because it wouldn’t need to in order to evaluate that new-if.